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3 years ago

As a cold air mass advances on a warm air mass, what usually comes before it?

1 answer:

8 0

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

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3 years ago

what is the density of a substance that has a mass of 2.0 g , and when placed in a graduated cylinder the volume changed from 70

lubasha [3.4K]

A material with a mass of 2.0 g when placed in a graduated cylinder the volume changed from 70 ml to 75 ml has a density of 0.4 g/mL.

How do I calculate the substance's density?

We'll start by getting the substance's volume. This is attainable as follows:

Water volume: 70 mL

75 mL = volume of material + water.

Substance volume =?

Substance volume equals (substance volume plus water) - (Volume of water)

Substance volume = 75 - 70

5 mL is the substance's volume.

Finally, we will calculate the substance's density. Below is an example to help:

2.0 g is the substance's mass.

5 mL is the substance's volume.

Substance density =?

Mass / volume equals density.

Substance density = 2/5

0.4 g/mL is the substance's density.

The density is therefore 0.4 g/mL.

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1 year ago

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

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3 years ago

A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.

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The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

$W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J$

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2 years ago

A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi

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Answer:

Explanation is given

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3 years ago

As a cold air mass advances on a warm air mass, what usually comes before it?​

As a cold air mass advances on a warm air mass, what usually comes before it?